How do you solve #x/(x-3)=4/(x-4)#?

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Answer 1 · 2016-09-07T00:34:05

#x = 6# and #x = 2#.

Use the property #a/b = m/n -> a xx n =bxxm#

#x(x - 4) = 4(x - 3)#

#x^2 - 4x = 4x - 12#

#x^2 - 8x + 12 = 0#

#(x - 6)(x - 2) = 0#

#x = 6 and 2#

Neither are extraneous, since the restrictions are #x!= 3# and #x!=4#, and the solutions conflict with neither.

Hopefully this helps!